Okay, so I have given this problem A shot. I got the answer so col$(A)$ however, I was just wondering if I did it correctly? Here is my work:
$$ A= \begin{bmatrix} 1 & 1 & -3 \\ 0 & 2 & 1 \\ 1 & -1 & -4 \\ \end{bmatrix} $$
and
$$ b= \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix} $$
I made an augmented matrix and reduced it all the way down to
$$ \left[ \begin{array}{ccc|c} 1&0&-7/2&1/2\\ 0&1&1/2&1/2\\ 0&0&0&0 \end{array} \right] $$
From this I made equations. I let
$x_3= t$.
I solved and got
$x2= 1/2 - 1/(2t)$
$x_1= 1/2 + 7/2(t)$
So, in the equation $Ax=b$. I got that
$$ X= \begin{bmatrix} 1/2 + 7/2(t) \\ 1/2 - 1/2(t) \\ t\\ \end{bmatrix} $$
And so, $b$ is in col$(A)$. Is this the correct way to solve this? I also, had another question. Now, I need to find out whether $w$ is in row$(A)$.
$$ w= \begin{bmatrix} 2 & 4 & -5 \\ \end{bmatrix} $$
I once again made it into an augmented matrix (except its a row... sorry I have no idea how to make that kind of matrix on here). I reduced it down but now I'm stuck and I think it is mainly because I'm not used to an augmented matrix looking like this. Anyway, this is what I get. The line is between row 3 and row 4.
$$ \begin{bmatrix} 1 & 0 & -7/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix} $$
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$\begingroup$I think this way is easier and more general:
To find the column space of a matrix, you don't need to use an augmented matrix. Just use column reduction, which will get you to: $$\left[ \begin{array}{ccc} 1&1&-3\\ 0&2&1\\ 1&-1&-4 \end{array} \right] \sim \left[ \begin{array}{ccc} 1&0&0\\ 0&2&1\\ 1&-2&-1 \end{array} \right] \sim \left[ \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 1&-1&0 \end{array} \right]$$ Then your column space is just the span of the nonzero column vectors. So: $$\text{span} \left(\begin{bmatrix} 1 \\ 0 \\ 1\\ \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -1\\ \end{bmatrix}\right)$$ Then to check if $b$ is in this span, just see if it can be represented as a linear combination of these 2 vectors. In this case $b = 1\begin{bmatrix} 1 \\ 0 \\ 1\\ \end{bmatrix} + 1\begin{bmatrix} 0 \\ 1 \\ -1\\ \end{bmatrix}$.
To find the row space, use row reduction instead of column reduction and then the nonzero rows will span your row space.
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